We said that the Laplace transformation of a product is not the product of the transforms. All hope is not lost however. We simply have to use a different type of a “product.” Take two functions \(f(t)\) and \(g(t)\) defined for \(t \geq 0\), and define the convolution3 of \(f(t)\) and \(g(t)\) as

Note the \(t\) in front of the sine. The solution, therefore, grows without bound as \(t\) gets large, meaning we get resonance.

Similarly, we can solve any constant coeﬃcient equation with an arbitrary forcing function \(f(t)\) as a deﬁnite integral using convolution. A deﬁnite integral, rather than a closed form solution, is usually enough for most practical purposes. It is not hard to numerically evaluate a deﬁnite integral.

3For those that have seen convolution deﬁned before, you may have seen it deﬁned as . This deﬁnition agrees with (6.2) if you deﬁne and to be zero for . When discussing the Laplace transform the deﬁnition we gave is suﬃcient. Convolution does occur in many other applications, however, where you may have to use the more general deﬁnition with inﬁnities.

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